Tuesday, November 1 ∗ Solutions ∗ Changing coordinates
1. Consider the region R in R2 shown below at right. In this problem, you will do a change of
coordinates to evaluate:
Ï
x − 2y dA
R
y
v
(3, 4)
T
(1, 3)
R
(1, 1)
(2, 1)
S
x
u
(a) Find a linear transformation T : R2 → R2 which takes the unit square S to R.
¡ ¢
Write you answer both as a matrix ac db and as T (u, v) = (au +bv, cu +d v), and check your
answer with the instructor.
SOLUTION:
T (u, v) = (2u + v, u + 3v). In matrix form,
Ã
2 1
1 3
!
Î
(b) Compute R x − 2y dA by relating it to an integral over S and evaluating that. Check your
answer with the instructor.
SOLUTION:
The Jacobian of T is
Ã
d et
So
Ï
2 1
1 3
!
= 6−1 = 5
Ï
R
x − 2y dA =
Z 1Z
=
0
1
0
S
[(2u + v) − 2(u + 3v)]5 dA
−25v 2
−25v d u d v =
2
·
¸1
= −25/2
0
2. Another simple type of transformation T : R2 → R2 is a translation, which has the general form
T (u, v) = (u + a, v + b) for a fixed a and b.
(a) If T is a translation, what is its Jacobian matrix? How does it distort area?
SOLUTION:
If T (u, v) = (u + a, v + b) where a and b are constants, then the Jacobian is
Ã
d et
1 0
0 1
!
= 1.
So T does not distort areas.
©
ª
(b) Consider the region S = u 2 + v 2 ≤ 1 in R2 with coordinates (u, v), and the region R =
©
ª
(x − 2)2 + (y − 1)2 ≤ 1 in R2 with coordinates (x, y).
Make separate sketches of S and R.
SOLUTION:
(c) Find a translation T where T (S) = R.
SOLUTION:
T (u, v) = (u + 2, v + 1)
(d) Use T to reduce
Ï
x dA
R
to an integral over S, and then evaluate that new integral using polar coordinates.
SOLUTION:
The Jacobian of T is just 1, as noted in part (a). So we have
Ï
Ï
R
x dA =
S
(u + 2) dA
Converting the second integral above to polar we have
Ï
2π Z 1
Z
S
(u + 2) dA =
0
Z
= 1/3
0
0
2π
(r cos θ + 2)r d r d θ =
Z
0
2π · r 3 cos θ ¸1
3
0
£ ¤1
d θ + 2π r 2 0
cos θ d θ + 2π = 1/3 [sin θ]2π
0 + 2π = 2π
3. Consider the region R shown below. Here the curved left side is given by x = y − y 2 . In this
problem, you will find a transformation T : R2 → R2 which takes the unit square S = [0, 1]×[0, 1]
to R.
y
(2, 1)
(0, 1)
R
x
(2, 0)
(a) As a warm up, find a transformation that takes S to the rectangle [0, 2] × [0, 1] which contains R.
SOLUTION:
L(u, v) = (2u, v)
(b) Returning to the problem of finding T taking S to R, come up with formulas for T (u, 0),
T (u, 1), T (0, v), and T (1, v). Hint: For three of these, use your answer in part (a).
SOLUTION:
T (u, 0) = (2u, 0) T (u, 1) = (2u, 1)
T (1, v) = (2, v)
T (0, v) = (v − v 2 , v)
(c) Now extend your answer in (b) to the needed transformation T . Hint: Try “filling in” between T (0, v) and T (1, v) with a straight line.
SOLUTION:
T (u, v) = (2u + v(1 − v)(1 − u), v)
(d) Compute the area of R in two ways, once using T to change coordinates and once directly.
SOLUTION:
To change coordinates we compute the Jacobian
Ã
J (T ) = d et
2 − v(1 − v) (1 − 2v)(1 − u)
0
1
!
= 2 − v(1 − v)
So we have the area of R given by
Z 1Z
Ï
R
dx dy =
0
1
0
2 − v(1 − v) d u d v = 11/6
Computing directly we have the area of R given by
Z
0
1
2 − (y − y 2 ) d y = 11/6
4. If you get this far, evaluate the integrals in Problems 1 and 2 directly, without doing a change of
coordinates. It’s a fun-filled task. . .
SOLUTION:
For the integral in problem one, use the order d y d x. We need to split the double integral into
three parts. The result is
Z 1Z
Ï
R
x − 2y dA =
0
3x
x/2
Z 2Z
x − 2y d y d x +
x/2+5/2
x − 2y d y d x +
x/2
1
Z 3Z
2
x/2+5/2
3x−5
x − 2y d y d x
Evaluating this is not difficult but it is tedious. We leave it to the interested student. You should
get −25/2.
For the integral in problem two, again use the order d y d x. We just need one double integral.
Z 3Z
Ï
R
x dA =
Z
=
1−(x−2)2
p
1−
1
3
2x
1
p
1+
1−(x−2)2
x dy dx
p
1 − (x − 2)2 d x
This integral can be evaluated by making the substitution x − 2 = sin u, yielding the integral
Z
π/2
−π/2
(2 sin u + 4) cos2 u d u
Now split this in two pieces as
Z
π/2
−π/2
2
2 sin u cos u d u +
Z
π/2
4 cos2 u d u
−π/2
The first is the integral of an odd function over an interval which is symmetric about the y axis
so it is 0. The second can be evaluated by using the trig identity cos2 u = (1 + cos 2u)/2. This
gives
Z π/2
Z π/2
2
4 cos u d u =
4(1 + cos 2u)/2 d u = 2π.
−π/2
−π/2